Alkyne Synthesis Reaction Problem | Summary and Q&A

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March 31, 2023
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Alkyne Synthesis Reaction Problem

TL;DR

The reaction between propine, sodium amide, and ethyl bromide forms 2-pentyne through a series of steps.

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Questions & Answers

Q: What is propine and how does it react with sodium amide and ethyl bromide?

Propine is a three-carbon alkyne. It reacts with sodium amide, which acts as a strong base and deprotonates the alkyne, forming an alkaline ion. This alkaline ion then undergoes an SN2 reaction with ethyl bromide, resulting in the formation of 2-pentyne.

Q: What is the role of sodium amide in the reaction?

Sodium amide acts as a strong base in the reaction. It deprotonates the alkyne CH hydrogen, forming an alkaline ion. This alkaline ion then acts as a nucleophile in the subsequent SN2 reaction.

Q: What is the final product of the reaction?

The final product of the reaction is 2-pentyne. It is formed through the SN2 reaction between the alkaline ion and the primary carbon of ethyl bromide.

Q: Why is the answer choice B, 2-pentyne?

The reaction of propine with sodium amide followed by ethyl bromide leads to the formation of a five-carbon system with a triple bond between carbons two and three. Therefore, the correct answer is 2-pentyne.

Summary & Key Takeaways

  • Propine, a three-carbon alkyne, reacts with sodium amide nanh2 followed by ethyl bromide ch3ch2 br.

  • Sodium amide acts as a strong base, deprotonates the alkyne CH hydrogen, and forms an alkaline ion.

  • The resulting alkaline ion behaves as a nucleophile and undergoes an SN2 reaction with the primary carbon of ethyl bromide, leading to the formation of 2-pentyne.

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